4.3 Power Series

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We determine radius of convergence of complex power series.

A complex power series centered at $\bf z_0$ is a complex series of the form $\sum _{k=0}^\infty c_k(z-z_0)^k = c_0 + c_1 (z-z_0) + c_2 (z-z_0)^2 + c_3 (z-z_0)^3 + \cdots$

Since a complex power series contains a variable, the question is, for which values of the variable, $z$ , does the complex power series converge? This question is answered by using the root test, as described in the following theorem.

Let $\dis L = \limsup _{k \to 0} \sqrt [k]{|c_k|}$ and let $\dis \rho = \frac {1}{L}$ . Then complex power series centered at $z_0$ given by $\sum _{k=0}^\infty c_k(z-z_0)^k$ will converge for $z \in D(z_0, \rho )$ and diverge for $|z-z_0| > \rho$ .
The radius, $\rho$ , is called the radius of convergence of the complex power series.
If $L = 0$ , then $\rho$ is understood to be infinity, i.e., the complex power series converges for all $z$ .
If $L = \infty$ , then the power series converges only at its center, $z_0$ .
A complex power series may converge or diverge points on the circle $C(z_0, \rho )$ .

The limit $L$ , can also be computed from the ratio test as $L = \limsup _{k \to \infty } \left |\frac {c_{k+1}}{c_k}\right |$ provided $c_k \neq 0$ for all but finitely many values of $k$

Proof: From the root test, the series $\dis \sum _{k=0}^\infty c_k(z-z_0)^k$ converges if $z$ satisfies the inequality $\limsup _{k \to 0} \sqrt [k]{|c_k(z-z_0)^k|} = |z-z_0| \limsup _{k \to 0} \sqrt [k]{|c_k|} = |z-z_0| \cdot L < 1$ If $L =0$ , then the series converges for all $z$ . If $L \neq 0$ , then the series converges if $|z-z_0| < \frac {1}{L} = \rho$ . Moreover, the root test also shows that if $|z-z_0| > \frac {1}{L} = \rho$ then the series diverges. $\blacksquare$

example 1 Find the disk of convergence of the complex power series: $\dis \sum _{k=1}^\infty \frac {i^k}{k}(z+i)^k$
We first compute the limit superior from the root test: $L = \limsup _{k \to \infty } \sqrt [k]{|c_k|} = \limsup _{k \to \infty } \sqrt [k]{\left |\frac {i^k}{k}\right |} = \limsup _{k \to \infty } \frac {1}{\sqrt [k]{k}} = 1$ Therefore the radius of convergence is $\rho = 1/L = 1$ and since the center of the power series is $-i$ , the disk of convergence is $D(-i, 1)$ .

(problem 1a) Find the disk of convergence of the complex power series: $\dis \sum _{k=1}^\infty \frac {e^{ik}}{k^2}(z-2i)^k$
The center of the power series is $\; \answer {2i}$
The limit superior of the coefficients from the root test (or ratio test) is: $\limsup _{k \to \infty } \sqrt [k]{|c_k|} = \; \answer {1}$ The disk of convergence is:

$D(-2i, 1)$
$D(2i, -1)$
$D(2i, 1)$

Here is a video solution to problem 1a:

(problem 1b) Find the disk of convergence of the complex power series: $\dis \sum _{k=1}^\infty \frac {k^3}{(1+i)^k}(z-1+3i)^k$
The center of the power series is $\; \answer {1-3i}$
The limit superior of the coefficients from the root test (or ratio test) is: $\limsup _{k \to \infty } \sqrt [k]{|c_k|} = \; \answer {2^{-1/2}}$ The disk of convergence is:

$D(1-3i, 1/\sqrt {2})$
$D(1-3i, \sqrt 2)$
$D(1-3i, 2)$

On its disk of convergence, a power series is an analytic function.

Suppose the complex power series $f(z) = \sum _{k=0}^\infty c_k(z-z_0)^k$ converges on $D(z_0, \rho )$ for some $\rho > 0$ . Then $f$ is differentiable on $D(z_0, \rho )$ and $f'(z) = \sum _{k=1}^\infty kc_k(z-z_0)^{k-1}$ Moreover, $f'$ has the same disk of convergence as $f$ .

The formula for $f'$ shows that a power series can be differentiated term by term.

Suppose the complex power series $f(z) = \sum _{k=0}^\infty c_k(z-z_0)^k$ converges on $D(z_0, \rho )$ for some $\rho > 0$ . Then $f$ is infinitely differentiable on $D(z_0, \rho )$

Proof: Since $f'$ is a power series with the same radius of convergence as $f$ , the theorem can be applied to $f'$ . Thus, $f'$ is differentiable and $f''$ has the same radius of convergence as $f'$ . Continuing inductively, $f$ has derivatives of all orders, i.e., $f$ is infinitely differentiable on $D(z_0, \rho )$ . $\blacksquare$

Suppose the complex power series $f(z) = \sum _{k=0}^\infty c_k(z-z_0)^k$ converges on $D(z_0, \rho )$ for some $\rho > 0$ . Then we have the following formula for the coefficients: $c_k = \frac {f^{(k)}(z_0)}{k!}$

Proof: Since $f$ is infinitely differentiable and the derivative is computed term by term, we have $f^{(j)}(z) = \sum _{k=j}^\infty k(k-1)(k-2)\cdots (k-j+1)c_k(z-z_0)^{k-j}$ Plugging in $z_0$ for $z$ and noting that all of the terms of the series are zero except the first (where $k = j$ ), we obtain $f^{(j)}(z_0) = j!c_j$ Note that we treat the zero power in the first term as 1 when $z=z_0$ . $\blacksquare$

example 2 Find the derivative of the power series function $\dis f(z) = \sum _{k=0}^\infty \frac {z^k}{k!}$
The ratio test shows that this series converges for all $z \in \C$ . It’s derivative is $f'(z) = \sum _{k=1}^\infty \frac {kz^{k-1}}{k!}= \sum _{k=1}^\infty \frac {z^{k-1}}{(k-1)!}$ Re-indexing so that $k$ starts at $0$ instead of $1$ (i.e. replacing $k-1$ with $k$ ), yields $f'(z) = \sum _{k=0}^\infty \frac {z^k}{k!}$ But this is the same series as the original function! What function is this?

(problem 2a) Find the derivative of the power series function $\dis f(z) = \sum _{k=0}^\infty z^k$
$\dis f'(z) = \sum _{k=0}^\infty \answer {(k+1)z^k}$

(problem 2b) Find the derivative of the power series function $\dis f(z) = \sum _{k=0}^\infty \frac {(-1)^k}{(2k+1)!}z^{2k+1}$
$\dis f'(z) = \sum _{k=0}^\infty \answer {\frac {(-1)^k}{(2k)!}z^{2k}}$

Here is a video solution to problem 2a:

2024-11-09 13:45:52